James Stewart-Calculus_ Early Transcendentals-Cengage Learning (2015)

478 Chapter 7 Techniques of Integration
68. A rocket accelerates by burning its onboard fuel, so its mass
decreases with time. Suppose the initial mass of the rocket
at liftoff (including its fuel) is m, the fuel is consumed at
rate r, and the exhaust gases are ejected with constant
velocity ve (rel ative to the rocket). A model for the velocity
of the rocket at time t is given by the equation
vstd − 2tt 2 ve ln m 2 rt
m
where t is the acceleration due to gravity and t is not too
large. If t − 9.8 mys 2 , m − 30,000 kg, r − 160 kgys, and
ve − 3000 mys, find the height of the rocket one minute
after liftoff.
69. A particle that moves along a straight line has velocity
vstd − t 2 e 2t meters per second after t seconds. How far will
it travel during the first t seconds?
70. If f s0d − ts0d − 0 and f 0 and t 0 are continuous, show that
y a
0
f sxdt0sxd dx − f sadt9sad 2 f 9sadtsad 1 y a
f 0sxdtsxd dx
71. Suppose that f s1d − 2, f s4d − 7, f 9s1d − 5, f 9s4d − 3,
and f 0 is continuous. Find the value of y 4 xf 0sxd dx.
1
72. (a) Use integration by parts to show that
y f sxd dx − xf sxd 2 y xf 9sxd dx
(b) If f and t are inverse functions and f 9 is continuous,
prove that
y b
a
f sxd dx − bf sbd 2 af sad 2 y f sbd
tsyd dy
[Hint: Use part (a) and make the substitution y − f sxd.]
(c) In the case where f and t are positive functions and
b . a . 0, draw a diagram to give a geometric interpretation
of part (b).
(d) Use part (b) to evaluate y e 1
ln x dx.
73. We arrived at Formula 6.3.2, V − y b a
2x f sxd dx, by using
cylindrical shells, but now we can use integration by parts
to prove it using the slicing method of Section 6.2, at least
f sad
0
for the case where f is one-to-one and therefore has an
inverse function t. Use the figure to show that
V − b 2 d 2 a 2 c 2 y d
ftsydg 2 dy
Make the substitution y − f sxd and then use integration by
parts on the resulting integral to prove that
V − y b
2x f sxd dx
74. Let I n − y y2
0
sin n x dx.
(a) Show that I 2n12 < I 2n11 < I 2n.
(b) Use Exercise 50 to show that
a
I 2n12
− 2n 1 1
I 2n 2n 1 2
(c) Use parts (a) and (b) to show that
2n 1 1
2n 1 2 < I2n11
I 2n
< 1
and deduce that lim n l ` I 2n11yI 2n − 1.
(d) Use part (c) and Exercises 49 and 50 to show that
2
lim
n l ` 1 ? 2 3 ? 4 3 ? 4 5 ? 6 5 ? 6 7 ? ∙ ∙ ∙ ? 2n
2n 2 1 ?
c
2n
2n 1 1 − 2
This formula is usually written as an infinite product:
2 − 2 1 ? 2 3 ? 4 3 ? 4 5 ? 6 5 ? 6 7 ? ∙ ∙ ∙
and is called the Wallis product.
(e) We construct rectangles as follows. Start with a square of
area 1 and attach rectangles of area 1 alternately beside or
on top of the previous rectangle (see the figure). Find the
limit of the ratios of width to height of these rectangles.
y
d
x=g(y)
y=ƒ
c
x=b
x=a
0 a
b x
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